Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k + 1}{k^2 + 19k + 90} \times \dfrac{2k^2 + 20k}{-9k - 9} $
Explanation: First factor the quadratic. $q = \dfrac{k + 1}{(k + 10)(k + 9)} \times \dfrac{2k^2 + 20k}{-9k - 9} $ Then factor out any other terms. $q = \dfrac{k + 1}{(k + 10)(k + 9)} \times \dfrac{2k(k + 10)}{-9(k + 1)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (k + 1) \times 2k(k + 10) } { (k + 10)(k + 9) \times -9(k + 1) } $ $q = \dfrac{ 2k(k + 1)(k + 10)}{ -9(k + 10)(k + 9)(k + 1)} $ Notice that $(k + 1)$ and $(k + 10)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 2k(k + 1)\cancel{(k + 10)}}{ -9\cancel{(k + 10)}(k + 9)(k + 1)} $ We are dividing by $k + 10$ , so $k + 10 \neq 0$ Therefore, $k \neq -10$ $q = \dfrac{ 2k\cancel{(k + 1)}\cancel{(k + 10)}}{ -9\cancel{(k + 10)}(k + 9)\cancel{(k + 1)}} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $q = \dfrac{2k}{-9(k + 9)} $ $q = \dfrac{-2k}{9(k + 9)} ; \space k \neq -10 ; \space k \neq -1 $